# F(y)=summation 1/(y^2+m^2) is not differentiable.

## Homework Statement

Is $$f(y)=\sum_{m=1}^\infty \frac{1}{y^2+m^2}$$differentiable?

## The Attempt at a Solution

From the graph, it is obvious that f is not differentiable at y=0, but I don't know how to prove that. I proved that $$\sum_{m=1}^n f_m=\sum_{m=1}^n\frac{1}{y^2+m^2}$$ converges uniformly to f. Does this help? Thank you very much.

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Dick
Homework Helper
The uniform convergence of differentiable functions does not prove the sum is differentiable. If the derivatives converge uniformly then it is. Don't you have a theorem like that? And why is it clear it's not differentiable at zero?

The uniform convergence of differentiable functions does not prove the sum is differentiable. If the derivatives converge uniformly then it is. Don't you have a theorem like that? And why is it clear it's not differentiable at zero?
I have been thinking that each $$f_m$$ is strictly increasing on $$(0,\infty)$$. Therefore, so is f(y). If f is differentiable, then f'(0)=0 which is not true. Is this correct?

Dick
Homework Helper
I have been thinking that each $$f_m$$ is strictly increasing on $$(0,\infty)$$. Therefore, so is f(y). If f is differentiable, then f'(0)=0 which is not true. Is this correct?
Do you mean decreasing? f'(0) has to be zero, the function is even. 1/(x^2+1) is strictly decreasing on [0,infinity). It's perfectly differentiable.

Do you mean decreasing? f'(0) has to be zero, the function is even. 1/(x^2+1) is strictly decreasing on [0,infinity). It's perfectly differentiable.
Yeah I meant decreasing.
I know that each $$f_m$$ is differentiable on R. But f is not. I don't know how to prove this though. I am trying to prove by contradiction, but I cannot find a contradiction.

Do you mean decreasing? f'(0) has to be zero, the function is even. 1/(x^2+1) is strictly decreasing on [0,infinity). It's perfectly differentiable.
I think that f is not differentiable at y=0 because the graph of f is really pointy at y=0...

Dick
Homework Helper
I think that f is not differentiable at y=0 because the graph of f is really pointy at y=0...
If f(y) looks really pointy at y=0 it might be a problem with the graphing program. This is an analysis course, right? You shouldn't rely on that.

If f(y) looks really pointy at y=0 it might be a problem with the graphing program. This is an analysis course, right? You shouldn't rely on that.
Hi,

Now I see how to show that it is differentiable at x=0, but still don't know how to show that it is differentiable on the rest of R. I have two tools: cauchy criterion and weierstrass m-test. I tried both but none worked. Can you give some hints? Thanks.

Dick